AMC 10 · 2022 · #16

Test the roots against the system they came from without re-solving anything. Sum: (3+√(2)) + (3-√(2)) = 6, as required. Product: (3+√(2))(3-√(2)) = 9 - 2 = 7, also as required. So the pair satisfies b+c=6 and bc=7 exactly, with no rounding involved. The numerical rebuild in Step 7 is the stronger check, because it never touches the quadratic at all: it starts from the claimed value of log₂ y, reconstructs x and y as actual decimals, and confirms x^y = 2⁶⁴ and (log₂ x)^log₂ y = 2⁷ to five decimal places for both candidates. Size check against the answer list: √(2) ≈ 1.41421, so 3+√(2) ≈ 4.41421, which lands between (B) 4 and (D) 4+√(3) ≈ 5.73205, a comfortable place for it to sit. Now audit the decoys, because each is a specific mistake. Choice (A) 3 is the midpoint 6/2 of the two roots, which is what you get by noticing the symmetry but dropping the ±√(2) spread, or by assuming b=c in b+c=6. Choice (B) 4 is 3+√(2) with √(2) carelessly rounded to 1, and it is also the value a guesser lands on when the radical answers look too exotic. Choice (E) 7 is the product bc rather than the root, and it is also the exponent sitting on the right of a^b = 2⁷, so it catches anyone who reads that line as b = 7. Choice (D) 4+√(3) is the correct answer to the same problem with 2²⁵⁶ and 2¹³ in place of 2⁶⁴ and 2⁷, since those give sum 8 and product 13, whose roots are 4 ± √(3); it is the shape of the right method applied to the wrong numbers. Finally, a consistency check on the domain work from Step 2: in both candidates c ＞ 0, so a = 2^c ＞ 1 and x = 2^a ＞ 2, comfortably positive, and y = 2^b ＞ 0 automatically. Nothing had to be thrown away for being out of range, which is precisely why the problem needed the word "greatest".